Question

The temperature coefficient for the saponification of ethyl acetate by $NaOH$ is $1.75$. Calculate the activation energy.

Answer 1

Given : $\frac{K_2}{K_1}=1.75$

$T_1={25}^{\circ }C=25+273=298K,T_2={35}^{\circ }C=35+273=308K$

(Since temperature coefficient is ratio of rate constants at ${35}^{\circ }C$ and ${25}^{\circ }C$ respectively.)

$\because 2.303\log \frac{K_2}{K_1}=\frac{E_a}{R}\left[\frac{T_2-T_1}{T_1{T}_2}\right]$

$\therefore 2.303\log 1.75=\frac{E_a}{1.987}\left[\frac{308-298}{308\times 298}\right]$

$E_a=10.207kcalmo{l}^{-1}$.